如何从Trie中检索给定长度的随机单词

Question

我有一个简单的Trie，我用它来存储大约80k个长度为2-15的单词。它非常适合检查字符串是否是单词；然而，现在我需要一种获得给定长度的随机单词的方法。换句话说，我需要“getRandomWord(5)”来返回一个5个字母的单词，所有5个字母的单词都有相同的机会被返回。

我能想到的唯一方法是选择一个随机数并遍历树的宽度--首先，直到我通过了所需长度的那么多单词。有没有更好的办法做到这一点？

可能没必要，但这是我的trie代码。

class TrieNode {
    private TrieNode[] c;
    private Boolean end = false;

    public TrieNode() {
        c = new TrieNode[26]; 
    }

    protected void insert(String word) {
        int n = word.charAt(0) - 'A';
        if (c[n] == null)
            c[n] = new TrieNode();
        if (word.length() > 1) {
            c[n].insert(word.substring(1));
        } else {
            c[n].end = true;
        }
    }

    public Boolean isThisAWord(String word) {
        if (word.length() == 0)
            return false;
        int n = word.charAt(0) - 'A';
        if (c[n] != null && word.length() > 1)
            return c[n].isThisAWord(word.substring(1));
        else if (c[n] != null && c[n].end && word.length() == 1)
            return true;
        else
            return false;
    }
}

class TrieBranch {
    TrieNode node;
    int letter;
    int depth;
    public TrieBranch(TrieNode n, int l, int d) {
        letter = l; node = n; depth = d;
    }
}

class Dict {

    final static int maxWordLength = 13;    
    final static int lettersInAlphabet = 26;
    TrieNode trie;
    int lengthFrequencyByLetter[][];
    int totalLengthFrequency[];

    public Dict() {
        trie = new TrieNode();
        lengthFrequencyByLetter = new int[lettersInAlphabet][maxWordLength + 1];
        totalLengthFrequency = new int[maxWordLength + 1];
    }

    public String getRandomWord(int length) {
        // Returns a random word of the specified length from the trie
        // First, pick a random number from 0 to [number of words with this length]
        Random r = new Random();
        int wordIndex = r.nextInt(totalLengthFrequency[length]);

        // figure out what the first letter of this word would be
        int firstLetter = -1, totalSoFar = 0;
        while (totalSoFar <= wordIndex) {
            firstLetter++;
            totalSoFar += lengthFrequencyByLetter[firstLetter][length];
        }
        wordIndex -= (totalSoFar - lengthFrequencyByLetter[firstLetter][length]);

        // traverse the (firstLetter)'th node of trie depth-first to find the word (wordIndex)'th word
        int[] result = new int[length + 1];
        Stack<TrieBranch> stack = new Stack<TrieBranch>();
        stack.push(new TrieBranch(trie.getBranch(firstLetter), firstLetter, 1));
        while (!stack.isEmpty()) {
            TrieBranch n = stack.pop();
            result[n.depth] = n.letter;

            // examine the current node
            if (n.depth == length && n.node.isEnd()) {
                wordIndex--;
                if (wordIndex < 0) {
                    // search is over
                    String sResult = "";
                    for (int i = 1; i <= length; i++) {
                        sResult += (char)(result[i] + 'a');
                    }
                    return sResult;
                }
            }

            // handle child nodes unless they're deeper than target length
            if (n.depth < length) {
                for (int i = 25; i >= 0; i--) {
                    if (n.node.getBranch(i) != null)
                        stack.push(new TrieBranch(n.node.getBranch(i), i, n.depth + 1));
                }
            }
        }
        return "failure of some sort";
    }
}

Answer 1

为了确保你有一个均匀的机会得到每个5个字母的单词，你需要知道有多少5个字母的单词在你的树。因此，在构建树时，将单词的长度添加到两个计数器中：总频率计数器和字母频率计数器：

int lengthFrequencyByLetter[letterIndex][maxWordLength-1]
int totalLengthFrequency[maxWordLength-1]

如果你有4000个5个字母的单词，其中213个以“d”开头，那么

lengthFrequencyByLetter[3][4] = 213

和

totalLengthFrequency[4] = 4000

在将所有内容添加到树之后。（字母“A”是0，“B”是1，……“Z”是25。）

从这里，您可以搜索给定长度的n的第1个单词，其中n是从均匀随机分布中挑选的随机整数，范围为(0,totallengthfrequence[length-1])。

lengthFrequencyByLetter[0][4]
lengthFrequencyByLetter[1][4]
lengthFrequencyByLetter[2][4]
lengthFrequencyByLetter[3][4]